This "Medium/Hard" problem involves an array of length n (where n is even) and an integer limit. You want to make the array "complementary." An array is complementary if for all i from 0 to n/2 - 1, the sum nums[i] + nums[n - 1 - i] is equal to the same constant S.
You can replace any element in the array with any integer between 1 and limit. Each replacement counts as one move. Your goal is to find the minimum moves to make the array complementary for some S.
Companies like CureFit use this to test a candidate's ability to use Difference Arrays or Line Sweep algorithms. It requires:
(a, b), what range of S requires 0, 1, or 2 moves.S would be , which is too slow.The pattern is Difference Array (Line Sweep).
For each pair (a, b):
S in the range [2, 2 * limit].S in the range [min(a, b) + 1, max(a, b) + limit]. (This "overwrites" one of the 2-move costs).S == a + b. (This "overwrites" one of the 1-move costs).By using a difference array to track the "change in moves" at specific boundaries, we can calculate the total moves for all possible S in one linear pass.
limit = 6, Pair (3, 4)
min(3,4)+1=4 to max(3,4)+6=10. In the range [4, 10], we only need 1 move.3+4=7. At S=7, we need 0 moves.
We repeat this for all pairs and find the S that has the lowest total moves.S from 2 to 2 * limit and iterating through all pairs for each.delta values at the boundaries of the ranges.The "Difference Array" is a powerful tool for problems involving "adding a value to a range." Whenever you have many ranges and you want to find a point that is covered by the most (or least) cost, think Line Sweep.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Contiguous Array | Medium | Solve | |
| Count of Interesting Subarrays | Medium | Solve | |
| Intervals Between Identical Elements | Medium | Solve | |
| Make Sum Divisible by P | Medium | Solve | |
| Maximum Good Subarray Sum | Medium | Solve |